This exercises is taken from:
Oliver Blanchard (2017) Macroeconomics (7 Edition) > Chapter 4 Financial Markets I > Questions and Problems > Exercise 2
Suppose that the houselhold nominal income for an economy, \(Y_t\), is \(\$50,000\) billions and the nominal demand for money, \(M_t^d\), in this economy is given by:
\[M_t^d = Y_t*(0.2 - 0.8i_t)\]
Where \(i_t\) is the nominal interest rate fixed by the central bank.
\[\left. M_t^d \right|_{i_t = 0.01,Y_t=50000} = 50000*(0.2-0.8*0.01) = 9600\] \[\left. M_t^d \right|_{i_t = 0.05,Y_t=50000} = 50000*(0.2-0.8*0.05) = 8000\] Where \(M_t^d\) is expressed in billions.
Possible answer 1:
New nominal income: \[\begin{split}Y_t & = 50000 - 50000*0.2 \\ & = 40000 \end{split}\]
New demand money: \[\begin{split} \left. M_t^d \right|_{Y_t=40000} = 40000*(0.2-0.8*i_t) \end{split}\]
Therefore \(\left. M_t^d \right|_{Y_t=50000} > \left. M_t^d \right|_{Y_t=40000}\) and this means that \(M_t^d\) also declines
\[\begin{split} \epsilon_{M_t^d,Y_t} & = \frac{dM_t^d}{dY_t}*\frac{Y_t}{M_t^d} \\ & = (0.2-0.8*i_t)\frac{Y_t}{Y_t*(0.2-0.8*i_t)} \\ & = 1 > 0 \end{split}\]
Where we are assuming that \((0.2-0.8*i_t) \neq 0\) and \(Y_t \neq 0\). If this is true then \(\epsilon_{M_t^d,Y_t} = 1\) means that if \(Y_t\) increases by \(20\%\) then \(M_t^d\) also increases by \(20\%\)
The demand of money, \(M_t^d\), depends positively on nominal income, \(Y_t\):
\[\frac{dM_t^d}{dY_t} = (0.2−0.8i_t) > 0 \] If \(0.25 > i_t\). But this is true because \(M_t^d\) can’t be negative so \(M_t^d > 0\) and this implies that \(Y_t*(0.2−0.8i_t) > 0\) where \(Y_t > 0\) so \((0.2−0.8i_t) > 0\).
This means that if nominal income increases then people tend to demand more money. This occurs because people want to purchase more products and this implies that they need to have more money in their pockets.
Consider, for example, the colombian case where some workers receive an additional income, known as “prima”, at the end of the year where individuals use this additional income to buy more products than usual and they need more money in their pockets to do that.
The demand of money, \(M_t^d\), depends negatively on the interest rate, \(i_t\):
\[\frac{dM_t^d}{di_t} = -0.8*Y_t < 0 \] Because \(Y_t > 0\). This means that if the interest rate increases then people tend to demand less money. This occurs because people have the incentives to save more. Therefore, they tend to give their money to financial intermediaries and will have less money in their pockets.
The central bank should decrease the interest rate so there will exist fewer incentives to save. In that way people will have the desire to have more money in their pockets. This situation will generate an increase in the amount of money demanded.
In the appendix of Chapter 4 Financial Markets I from the book Oliver Blanchard (2017) Macroeconomics (7 Edition) there is an explanation about the demand for central bank money.
http://www.banrep.gov.co/ > Estadísticas > Tasas de interés y sector financiero > 3. Agregados monetarios, encaje y cartera > Agregados monetarios > Descargar y consultar: Efectivo, Base monetaria, M3 y sus componentes - Mensual > Exportar
Section 4-3 Determining the interest rate II mentions the federals fund rate as the \(i_t\) for the model in the USA. In Colombia \(i_t\) is know as “Tasa de intervención del Banco de la República”.
http://www.banrep.gov.co/ > Estadísticas > Tasas de interés y sector financiero > 1. Tasas de interés de política monetaria > Tasas de interés de política monetaria > Descargar y consultar: Serie diaria (desde 04/01/1999)
<https://databank.worldbank.org/source/world-development-indicators>
This exercises is based on:
Oliver Blanchard (2017) Macroeconomics (7 Edition) > Chapter 5 Goods and Financial Markets; The IS-LM model > Questions and Problems > Exercise 5
Consider the following numerical example of the IS-LM model:
\[C_t = 100 + 0.3Y_{tD}\]
\[I_t = 150 + 0.2Y_t - 1000i_t\] \[T_t = 100\] \[G_t = 200\] \[\overline{i}_t = 0.01 (1\%)\] \[(\frac{M}{P})^d = 2Y_t - 4000i_t\]
\[\begin{split} Z_t & \equiv C_t + I_t + G_t \\ Z_t & \equiv 100 + 0.3Y_{tD} + 150 + 0.2Y_t - 1000i_t + G_t \\ Z_t & \equiv 100 + 0.3(Y_t - 100) + 150 + 0.2Y_t - 1000i_t + 200 \\ Z_t & \equiv 450 + 0.3(Y_t - 100) + 0.2Y_t - 1000i_t \\ Z_t & \equiv 420 + 0.5Y_t - 1000i_t \end{split}\]
\[\begin{split} Y_t & = Z_t \\ Y_t & = 420 + 0.5Y_t - 1000i_t \\ Y_t & = \frac{420 - 1000i_t}{0.5} \\ Y_t & = 840 - 2000i_t \end{split}\]
\[\begin{split} i_t & = \overline{i_t} \\ i_t & = 0.01 \end{split}\]
\[\begin{split} Y_t & = 840 − 2000*0.01 \\ Y_t & = 820 \end{split}\]
\[i_t= 0.01\]
\[\begin{split} C_t & = 100 + 0.3(Y_t - 100) \\ C_t & = 100 + 0.3(820 - 100) \\ C_t & = 316 \end{split}\]
\[\begin{split} I_t & = 150 + 0.2*820 - 1000*0.01 \\ I_t & = 304 \end{split}\]
\[\begin{split} (\frac{M}{P})^s & = (\frac{M}{P})^d \\ (\frac{M}{P})^s & = 2Y_t - 4000i_t \\ (\frac{M}{P})^s & = 2*820 - 4000*0.01 \\ (\frac{M}{P})^s & = 2*820 - 4000*0.01 \\ (\frac{M}{P})^s & = 1600 \end{split}\]
Initially, if the central bank fix the money supply, \((\frac{M}{P})^s = 1500\), then also the central bank must increase \(i_t\) so people will accept to have less money in their pockets. Therefore the LM will be affected taking into account that it is defined as \(i_t = \overline{i}_t\). This effect can be seen later when we solve the next point where \(\frac{di_t}{d(\frac{M}{P})^s} = -\frac{1}{4000} < 0\).
Also the IS curve will be affected because the decrease in the money supply, \((\frac{M}{P})^s\), implies a change in the interest rate, \(i_t\), that affects investment, \(I_t\). Because investment, \(I_t\) will change then the IS will be affected and the direction of the change will depend on how investment is affected because \(\frac{dIS_t}{dI_t} > 0\).
\[\begin{split} (\frac{M}{P})^s & = (\frac{M}{P})^d \\ (\frac{M}{P})^s & = 2Yt−4000i_t \\ 4000i_t & = 2Yt - (\frac{M}{P})^s \\ i_t & = \frac{2Yt - (\frac{M}{P})^s}{4000} \end{split}\] If \((\frac{M}{P})^s = 1500\) then:
\[i_t = \frac{2Yt - 1500}{4000}\] Replacing \(i_t\) into the IS curve:
\[\begin{split} Yt & = 840−2000(\frac{2Yt - 1500}{4000}) \\ Yt & = 840−(\frac{2Yt - 1500}{2}) \\ Yt & = 840 − (Yt - 750) \\ 2Yt & = 840 + 750 \\ Yt & = 795 \end{split}\]
\[\begin{split} i_t & = \frac{2*795 - 1500}{4000} \\ i_t & = 0.0225 \; (2.25\%) \end{split}\]
\[\begin{split} C_t & = 100 + 0.3(795 - 100) \\ C_t & = 308.5 \end{split}\]
\[\begin{split} I_t & = 150 + 0.2*795 - 1000*0.0225 \\ I_t & = 286.5 \end{split}\]